608. Tree Node
Description
Table: Tree
+-------------+------+ | Column Name | Type | +-------------+------+ | id | int | | p_id | int | +-------------+------+ id is the column with unique values for this table. Each row of this table contains information about the id of a node and the id of its parent node in a tree. The given structure is always a valid tree.
Each node in the tree can be one of three types:
- "Leaf": if the node is a leaf node.
- "Root": if the node is the root of the tree.
- "Inner": If the node is neither a leaf node nor a root node.
Write a solution to report the type of each node in the tree.
Return the result table in any order.
The result format is in the following example.
Example 1:
Input: Tree table: +----+------+ | id | p_id | +----+------+ | 1 | null | | 2 | 1 | | 3 | 1 | | 4 | 2 | | 5 | 2 | +----+------+ Output: +----+-------+ | id | type | +----+-------+ | 1 | Root | | 2 | Inner | | 3 | Leaf | | 4 | Leaf | | 5 | Leaf | +----+-------+ Explanation: Node 1 is the root node because its parent node is null and it has child nodes 2 and 3. Node 2 is an inner node because it has parent node 1 and child node 4 and 5. Nodes 3, 4, and 5 are leaf nodes because they have parent nodes and they do not have child nodes.
Example 2:
Input: Tree table: +----+------+ | id | p_id | +----+------+ | 1 | null | +----+------+ Output: +----+-------+ | id | type | +----+-------+ | 1 | Root | +----+-------+ Explanation: If there is only one node on the tree, you only need to output its root attributes.
Note: This question is the same as 3054: Binary Tree Nodes.
Solutions
Solution 1: Conditional Statements + Subquery
We can use the CASE WHEN
conditional statement to determine the type of each node as follows:
- If a node's
p_id
isNULL
, then it is a root node. - Otherwise, if a node is the parent node of another node (we use a subquery to determine this), then it is an internal node.
- Otherwise, it is a leaf node.
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